It is easy to see that $\det(S)=\pm1$. However, one can show that $\det(S)=+1$ as we will now show using elementary methods.

Assuming $\det(A)\neq0$ when $S$ is written as a block matrix, one has $\det(S)= \det(AD-ACA^{-1}B) =\det(A) \det(D- CA^{-1}B)$. Thus, we need to show that $\det(A^{-1}) =\det(D- CA^{-1}B)$.

The second condition (of the three conditions given above) implies that $\det(A^{-1})=\det(D^{\textrm{T}}-A^{-1} B C^{\textrm{T}})=\det(D- C B^{\textrm{T}}(A^{-1})^{\textrm{T}})$.

Now using the first condition, we see that $B^{\textrm{T}} (A^{-1})^{\textrm{T}}= A^{-1} B$. Hence $\det(D- C B^{\textrm{T}}(A^{-1})^{\textrm{T}}) = \det(D- CA^{-1}B)=\det(A^{-1})$.

This completes the proof that $\det(S)=+1$ under the assumption that $\det(A)\neq0$. The other cases can be dealt with in a similar fashion and we won't pursue them here. A more sophisticated proof makes use of the Pfaffian (of an antisymmetric matrix).